I know immediately from this that any $\sigma$-algebra $\mathcal{F}$ must, at the very least, contain elements of $\mathcal{F}_1$: $$\{\Omega, \emptyset, \{a\}\}$$
but other than literally listing every possible collection of subsets of $\Omega$ and checking which are $\sigma$-algebras, and then intersecting such sets, I don't see what's an efficient way to do this.

$\begingroup$Complete the collection of subsets to be closed under the conditions for an sigma algebra, i.e., closed under the operations of countable union or intersection and complementary. This is the minimal sigma algebra that contain the collection.$\endgroup$
– MasacrosoFeb 22 '16 at 19:26

$\begingroup$And we know that this $\{b, c, d\}$ set with everything else in $\mathcal{F}_1$ must be the "smallest" $\sigma$-algebra generated by $\mathcal{F}_1$. Thank you! ... now, the question is, how do I prove this rigorously? I don't see how else to do this other than finding every possible $\mathcal{F}$ and then intersecting them.$\endgroup$
– ClarinetistFeb 22 '16 at 19:31

1

$\begingroup$@Clarinetist, no, it's the smallest also because it is the minimum extension of the current set to a sigma-algebra. So suffices to find one, and show that none of its subsets are sigma algebras.$\endgroup$
– gt6989bFeb 22 '16 at 19:34

Here are some more general statements that imply the desired equation. We assume the following without proof.

The intersection of $\sigma$-algebras is again a $\sigma$-algebra.

A set of the form $\lbrace \emptyset, \Omega, A, A^c \rbrace$ is a $\sigma$-algebra.

Let $\mathscr{G}$ (particular case $\mathcal{F}_1$) be any set of subsets of $\Omega$. $\mathscr{A}:=\bigcap_{\mathcal{F} \in \mathcal{I}(\mathscr{G})}\mathcal{F}$ is nonempty because the powerset $\mathscr{P}(\Omega)$ has $\mathscr{P}(\Omega) \in \mathcal{I}(\mathscr{G})$. It follows that $\mathscr{G} \subset \mathscr{A}$ and that, because of assumption 1, $\mathscr{A}$ is a $\sigma$-algebra. If $\mathscr{A}'$ is another $\sigma$-algebra with $\mathscr{G} \subset \mathscr{A}'$, then, because $\mathscr{A}'$ is one of the sets over which the intersection is taken, we have $\mathscr{A} \subset \mathscr{A}'$. So $\mathscr{A}$ satisfies the definition of the smallest $\sigma$-algebra generated by $\mathscr{G}$, i.e. $\sigma(\mathscr{G}) = \mathscr{A}$.

It holds in general that $\sigma(\lbrace A \rbrace) = \sigma(\lbrace \emptyset, \Omega, A \rbrace) = \lbrace \emptyset, \Omega, A, A^c \rbrace$, we could have proved this instead of proving this for our particular case in the last step.